We can write the equation of the tangent line in point-slope form, which is y - y0 = m(x - x0), where (x0, y0) is the point given to us and m is the slope (or derivative dy/dx) at this point. We are given the point (-2, 3), so now we have y - 3 = m(x + 2).
To find the derivative at this point, we first have to find dy/dx of this function using implicit differentiation with respect to x. The work is shown below.
d/dx [ln(10 - y2) + 16] = d/dx [-2x3]
⇒ d/dx[10 - y2] * 1 / (10 - y2) = -6x2 (the left-hand side is the chain rule used on ln. Remember that d/dx [ln(x)] = 1/x.
⇒ -2y * dy/dx * 1 / (10 - y2) = -6x2 (The resulting dy/dx is also a result of the chain rule).
Solving the above equation for the slope dy/dx, we get
dy/dx = 3x2(10 - y2) / y.
To find the derivative at the given point (-2, 3), we plug in x = -2 and y = 3 into dy/dx, getting
dy/dx at (-2, 3) = m = 3(-2)2(10 - 32) / 3 = 4
Substituting this into the equation for the line, we have our answer of
y - 3 = 4(x + 2)
or in slope-intercept form,
y = 4x + 11